Practice Quiz with Solutions: Chapter 7 (20 min)

Disclaimer: This quiz is representative of the level of
difficulty you should expect, but it doesn’t include every single
topic from the week’s work. The real quiz may include some other
topics and may skip some that are in this practice quiz. (The real
quiz also may not word questions in the same way as the practice
quiz. You should focus on the concepts, not a particular form of
words.)

These solutions show about the same level of work I expect from
you,
though I often add some extra commentary.
Please see Show Your Work for
the what, why, and how.

Caution: Your TI-83 shows
the curve and shaded area correctly, but because of the limitations of
screen area the TI-83 graphs are not labeled correctly. Your
lower bound (x or z)
must always be under the left edge of the shaded area, and your
upper bound must always be under the right edge of the shaded
area. The area number must not be below the axis at all, but above and
next to the shaded area.

1(points: 3½) Suppose that the new Ford Behemoth SUV gets a mean of
6.20 miles to the gallon, normally distributed with standard
deviation 0.90 mpg. Jane, who has taken a statistics
class, found that 85% of Ford
Behemoths get better gas mileage than hers does. What is her car’s
gas mileage, to two decimal places?

Solution:
You should have a normal curve with μ = 6.2
and σ = 0.9. The right-hand 85% of the curve
should be shaded in.

invNorm(1–.85, 6.2, 0.9) = 5.2672 →
5.27 mpg

Common mistake:
Some students calculate
invNorm(0.85,...) instead of invNorm(1−.85,...). Remember that
invNorm wants area to left, but the 85% is area to
right.

To guard against mistakes like this, always ask
yourself whether your answer is reasonable.
Here you know that 85% of cars get more mileage
than hers. If 50% of cars got better mileage than hers, her car would
be average (6.2 mpg). But since more than 50% of cars get better
mileage than hers, her car’s mileage must be below average, less than
6.2. An answer greater than 6.2 should ring your alarm bells.

2(points: 1) What is the percentile rank of Jane’s car?

Solution:
If 85% of cars get better mileage than hers, then 15% get
worse mileage. Her car is at the 15th %ile.

3(points: 3½) The times to assemble a particular product part are
normally distributed with a mean of 47.5 minutes and a standard
deviation of 8.5 minutes. What percent of assembly workers require
more than 60 minutes?

Solution:
The problem asks for P(x>60).
You should have a normal curve with μ = 47.5 and
σ = 8.5 as given above. 60 minutes is between one
and two s.d. above the mean; therefore the shaded
tail will be of moderate size. You could do ShadeNorm (as shown at
right) and read off the answer, 0.0707 or 7.07%. Or you could use
normalcdf:

P(x>60) = normalcdf(60, 10^99, 47.5, 8.5) =
.0707 = 7.07%

4(points: 1½) The standard normal curve is plotted in terms
of _____ (write in “z” or “x”).
It has a mean of _____ and a standard
deviation of ___.

Answer:
The standard normal curve is plotted in terms
of z.
It has a mean of 0 and a standard
deviation of 1.

5(points: 3) The lifetimes of Everlast AA batteries are normally
distributed. Between x=28.2 and x=34.1 hours, the area under the curve is
0.0742. Give two interpretations of this fact, in English.

Answer:

There’s a 7.42% probability that a randomly selected
Everlast AA battery has a lifetime between 28.2 and 34.1 hours.

7.42% of Everlast AA batteries have lifetimes between 28.2 and
34.1 hours.

Remember: probability of one = proportion of all.
Sometimes one interpretation is more natural, sometimes the other.

6(points: 2½) Fourteen rats were sent into space, and on their return to
Earth their red-blood-cell masses were measured. Here are the
data, in milliliters:

8.59 6.87 7.00 6.39 7.43 9.79 9.30
8.64 7.89 8.80 7.54 7.21 6.85 8.03

Are those figures normally distributed? Support your answer,
but you need not draw a graph.

Solution:
Use MATH200A part 4,
the normality check. You’ll see a
duplicate of the screen shot at right. The points are very nearly a
straight line, confirmed by r=.9838 and CRIT=.9351.
r>CRIT, and therefore you can say that
the rats’ red-blood-cell masses are normally distributed.